I'm trying to free your mind, Neo. But I can only show you the door. You're the one that has to walk through it.
I'm trying to free your mind, Neo. But I can only show you the door. You're the one that has to walk through it.
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HOWTO /
MomentumResolvedESIntroductionIn this tutorial, we will calculate the momentum-resolved entanglement spectrum of a 2D model on an infinite cylinder. The bipartition ... notation T_A T_B for partitions A, B {$\vert \psi \rangle = \sum_{ab} \psi_{ab} \vert a \rangle \otimes \vert b \rangle$} The Schmidt decomposition is equivalent to the singular value decomposition (SVD) of the matrix {$\psi_{ab}$}. This says that we can apply separate unitary transformations to the bases {$\vert a \rangle$} and {$\vert b \rangle$} such that in the new basis, the matrix {$\psi'_{a'b'}$} is diagonal and non-negative. The positive coefficients in the diagonal matrix are often called Schmidt coefficients. An equivalent procedure is to form the so-called reduced density matrix, which is the density matrix obtained by tracing over one half of the system, here we assume that we trace over the B region. That is, define {$\rho_{a'a} = \sum_b \psi_{a'b} \psi_{ab}^*$} The eigenvalues of the reduced density matrix are simply the square of the Schmidt coefficients. The entanglement spectrum is a different way of interpreting the density matrix. It is defined by {$H_e = -\ln \rho$} where {$\rho$} is the reduced density matrix at some bipartition of the wavefunction. This allows us to interpret the density matrix as corresponding to the entanglement Hamiltonian at a finite temperature {$T=1$}, {$\rho = e^{-H_{e}}$} The entanglement spectrum inherits all of the global symmetries of the wavefunction, so that each eigenstate of
{$H_e$} can be labelled by quantum numbers denoting some irreducible representation. For quantum numbers that are explicitly preserved by the MPS, these are simply the labels of the corresponding basis states, that you can see when looking at the density matrix (for example, with ModelThe example that we will look at is the square lattice on an infinite cylinder. We choose to have the cylinder aligned with the lattice directions, so that the system has a translation symmetry of a rotation by {$2\pi/W$}, where {$W$} is the width of the cylinder. <figure> To put this into a form suitable for DMRG, we need to map this onto a 1D chain. We use the mapping: <figure> The example Hamiltonian that we will use is an anisotropic spin-1/2 Heisenberg model, representing weakly-coupled rings (this has relatively low entanglement so the calculation converges quickly). Obtaining the groundstate wavefunctionFirstly, we construct the lattice, using width y=8, and using {$SU(2)$} symmetry. spincylinder-su2 -y 8 -o lattice Now we run DMRG until convergence mp-idmrg-s3e SymmetriesWe can get a quick view of the entanglement spectrum with mp-info: mp-info -d -p 0 psi We can see that the eigenvalues are labelled by the total spin. However many of the eigenvalues have an additional 2-fold degeneracy. This is related to the spatial symmetry of the wavefunction. The wavefunction should also be an eigenstate of the transverse momentum, that is the momentum around the cylinder. The transverse momentum is related to the eigenvalue of the translation operator {$T$}, {$T \vert k \rangle = e^{ik} \vert k \rangle$} where {$T$} rotates the cylinder by 1 site. If the wavefunction is non-chiral (that is, it doesn't spontaneously break reflection symmetry) then every momentum {$k$} eigenstate of the entanglement Hamiltonian will have a degenerate partner with momentum {$-k$}, except if the momentum is {$0$} or {$\pi$} the state is its own partner and will in general be non-degenerate. Lets check and see how well the wavefunction preserves the transverse momentum symmetry. If this really is a symmetry of the wavefunction, then the overlap {$\langle \psi \vert T \vert \psi \rangle$} should be {$\pm 1$}. We can check this with mp-ioverlap --string lattice:T -q 0 psi psi Here, we specified the lattice:"prod_unit(swapb(0,0)[0,1]*swapb(0,0)[1,2]*swapb(0,0)[2,3]*\ swapb(0,0)[3,4]*swapb(0,0)[4,5]*swapb(0,0)[5,6]*swapb(0,0)[6,7])" If the calculation has gone well, then the eigenvalue of the T operator, shown by Momentum-resolved spectrumWe have now established that the transverse momentum is a global symmetry of the wavefunction. Now what happens to the entanglement spectrum under this symmetry? We have established that {$\vert \psi \rangle$} is an eigenstate: {$ T \vert \psi \rangle = \vert \psi \rangle$}. Now, we have constructed the model in such a way that the {$T$} operator is separable at the boundary of the wavefunction unit cell. That is, as long as we choose our bipartition to be at the unit cell boundary, then we can decompose the {$T$} operator in the same way as the wavefunction. {$\vert \psi \rangle = \sum_{ab} \psi_{ab} \vert a \rangle \otimes \vert b \rangle$} {$T = T_A \otimes T_B$} where {$T_A$} and {$T_B$} are both unitary operators. Then the reduced density matrix of {$T \vert \psi \rangle $} is {$T_A \rho T_A^\dagger$} But we know that this must be equal to {$\rho$} itself, since {$\vert \psi \rangle$} is an eigenstate of {$T$}, hence {$T_A$} must be a symmetry of {$\rho$}, that is, {$ [T_A, \rho] = 0 $} Hence we can label the eigenstates of {$\rho$} (or equivalently, the entanglement Hamiltonian) by the eigenvalues of {$T_A$}. Note: if we didn't select the unit cell properly, then the {$T$} operator wont be separable into a simple product {$T = T_A T_B$}. This requires that the MPO representation of {$T$} has a bond dimension of 1 at the point where we construct the bipartition. In our case this is guaranteed because {$T$} only mixes sites within the unit cell, it doesn't contain any terms that cross the unit cell boundary. An equivalent statement is that if we coarse-grained the MPS and operator to put the 8-site unit cell into a single site, then {$T$} would become a product operator. But then we wouldn't need mp-ies psi lattice:T |